On Symmetry Through-put linking 3-4-5-6-7 Dimension's ,
Octo-Symmetry for the { 4-Dimension Quasi-Octahedron } ,
and 2 Identical { 8-D Structures } oriented Symmetrically

by Frank C. Fung ( 1st published in July, 2017. )

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Section IX - 24 Squares arising from the { 4-D Quasi-Cube }

Summary for the Section :

• In this Section , we shall identify :
  • the 24 Squares arising from the { 4-Dimension Quasi-Cube } .

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Recalling the 16 Position Vectors from the last Section :

• We simply bring-in again the 16 Position Vectors associated with :
  • the 16 { corners / nodal-points } of the { 4-Dimension Quasi-Cube } .

    Cross-Diagonal Line I.D.	* * *	Position Vector for 1st end-point of the Cross-Diagonal Line					* * *	Position Vector for 2nd end-point of the Cross-Diagonal Line
    	*	Vector I.D.	Co-ordinates				*	Vector I.D.	Co-ordinates
    *********************	*	***********	***	***	***	***	*	***********	***	***	***	***
    Cross-Diagonal Line 1	*	Vector A1	+1	+1	+1	+1	*	Vector A2	-1	-1	-1	-1
    Cross-Diagonal Line 2	*	Vector A3	+1	+1	-1	-1	*	Vector A4	-1	-1	+1	+1
    Cross-Diagonal Line 3	*	Vector A5	+1	-1	-1	+1	*	Vector A6	-1	+1	+1	-1
    Cross-Diagonal Line 4	*	Vector A7	+1	-1	+1	-1	*	Vector A8	-1	+1	-1	+1
    *********************	*	***********	***	***	***	***	*	***********	***	***	***	***
    Cross-Diagonal Line 5	*	Vector B1	+1	+1	+1	-1	*	Vector B2	-1	-1	-1	+1
    Cross-Diagonal Line 6	*	Vector B3	+1	+1	-1	+1	*	Vector B4	-1	-1	+1	-1
    Cross-Diagonal Line 7	*	Vector B5	+1	-1	+1	+1	*	Vector B6	-1	+1	-1	-1
    Cross-Diagonal Line 8	*	Vector B7	-1	+1	+1	+1	*	Vector B8	+1	-1	-1	-1

  And we simply take note here that :

  • each of the 16 Position Vectors is of { Length [ 2 ] } ,

    so that :

    • whenever 2 Position Vectors intersect one-another at an angle of [ 60 degrees ] ,
      • the line joining the 2 end-points of the 2 Position Vectors is also of { Length [ 2 ] } .

• Let us now identify the 32 pairs of Position Vectors intersecting an an angle of [ 60 degrees ] :
  • { Vector A1 / Vector B1 } , { Vector A1 / Vector B3 } , { Vector A1 / Vector B5 } , { Vector A1 , Vector B7 } ,
  • { Vector A3 / Vector B1 } , { Vector A3 / Vector B3 } , { Vector A4 / Vector B5 } , { Vector A4 , Vector B7 } ,
  • { Vector A5 / Vector B1 } , { Vector A6 / Vector B3 } , { Vector A5 / Vector B5 } , { Vector A6 , Vector B7 } ,
  • { Vector A8 / Vector B1 } , { Vector A7 / Vector B3 } , { Vector A7 / Vector B5 } , { Vector A8 , Vector B7 } ,

  • { Vector A2 / Vector B2 } , { Vector A2 / Vector B4 } , { Vector A2 / Vector B6 } , { Vector A2 , Vector B8 } ,
  • { Vector A4 / Vector B2 } , { Vector A4 / Vector B4 } , { Vector A3 / Vector B6 } , { Vector A3 , Vector B8 } ,
  • { Vector A6 / Vector B2 } , { Vector A5 / Vector B4 } , { Vector A6 / Vector B6 } , { Vector A5 , Vector B8 } ,
  • { Vector A7 / Vector B2 } , { Vector A8 / Vector B4 } , { Vector A8 / Vector B6 } , { Vector A7 , Vector B8 } .

  so that :

  • we now have 32 lines of { Length [ 2 ] } arising from joining the end-points of the { 32 pairs of Vectors } above .

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Setting up the 1st Square :

• Let us now take a look at these four (4) Lines , all of { Length [ 2 ] } :
  • { Line A1-B1 } ,
  • { Line B1-A3 } ,
  • { Line A3-B3 } and
  • { Line B3-A1 } ;

    so that the { 4 lines together } will form a { 4 nodal-points structure } .

  It is necessary for us to determine whether the { 4 nodal-points structure } is a square , a rhombus or a tetrahedron .

  And we see here that :

  • IF :
    • the 2 Cross-Diagonal Lines of the { 4 nodal-points structure } ,
      • i.e. : { Line A1-A3 } and { Line B1-B3 } in the case above ,
        • do intersect and intersect at their mid-points ;

    THEN :

    • the { 4 nodal-points structure } is a square .

• Let us now take a look at :
  • FIRST , the 24 mid-points arising from the 8 { Series-A Vectors } :

    Vector I.D.		Co-ordinates				*	Vector I.D.		Co-ordinates
    ***********************	*	***	***	***	***	*	***********************	*	***	***	***	***
    Vector A1		+1	+1	+1	+1	*	Vector A2		-1	-1	-1	-1
    Vector A3		+1	+1	-1	-1	*	Vector A4		-1	-1	+1	+1
    Vector A5		+1	-1	-1	+1	*	Vector A6		-1	+1	+1	-1
    Vector A7		+1	-1	+1	-1	*	Vector A8		-1	+1	-1	+1
    ***********************	*	***	***	***	***	*	***********************	*	***	***	***	***
    Mid-Point of Line A1-A3		+1	+1	0	0	*	Mid-Point of Line A2-A4		-1	-1	0	0
    Mid-Point of Line A1-A4		0	0	+1	+1	*	Mid-Point of Line A2-A3		0	0	-1	-1
    Mid-Point of Line A1-A5		+1	0	0	+1	*	Mid-Point of Line A2-A6		-1	0	0	-1
    Mid-Point of Line A1-A6		0	+1	+1	0	*	Mid-Point of Line A2-A5		0	-1	-1	0
    Mid-Point of Line A1-A7		+1	0	+1	0	*	Mid-Point of Line A2-A8		-1	0	-1	0
    Mid-Point of Line A1-A8		0	+1	0	+1	*	Mid-Point of Line A2-A7		0	-1	0	-1
    Mid-Point of Line A3-A5		+1	0	-1	0	*	Mid-Point of Line A4-A6		-1	0	+1	0
    Mid-Point of Line A3-A6		0	+1	0	-1	*	Mid-Point of Line A4-A5		0	-1	0	+1
    Mid-Point of Line A3-A7		+1	0	0	-1	*	Mid-Point of Line A4-A8		-1	0	0	+1
    Mid-Point of Line A3-A8		0	+1	-1	0	*	Mid-Point of Line A4-A7		0	-1	+1	0
    Mid-Point of Line A5-A7		+1	-1	0	0	*	Mid-Point of Line A6-A8		-1	+1	0	0
    Mid-Point of Line A5-A8		0	0	-1	+1	*	Mid-Point of Line A6-A7		0	0	+1	-1

  • SECONDLY , the 24 mid-points arising from the 8 { Series-B Vectors } :

    Vector I.D.		Co-ordinates				*	Vector I.D.		Co-ordinates
    ***********************	*	***	***	***	***	*	***********************	*	***	***	***	***
    Vector B1		+1	+1	+1	-1	*	Vector B2		-1	-1	-1	+1
    Vector B3		+1	+1	-1	+1	*	Vector B4		-1	-1	+1	-1
    Vector B5		+1	-1	+1	+1	*	Vector B6		-1	+1	-1	-1
    Vector B7		-1	+1	+1	+1	*	Vector B8		+1	-1	-1	-1
    ***********************	*	***	***	***	***	*	***********************	*	***	***	***	***
    Mid-Point of Line B1-B3		+1	+1	0	0	*	Mid-Point of Line B2-B4		-1	-1	0	0
    Mid-Point of Line B1-B4		0	0	+1	-1	*	Mid-Point of Line B2-B3		0	0	-1	+1
    Mid-Point of Line B1-B5		+1	0	+1	0	*	Mid-Point of Line B2-B6		-1	0	-1	0
    Mid-Point of Line B1-B6		0	+1	0	-1	*	Mid-Point of Line B2-B5		0	-1	0	+1
    Mid-Point of Line B1-B7		0	+1	+1	0	*	Mid-Point of Line B2-B8		0	-1	-1	0
    Mid-Point of Line B1-B8		+1	0	0	-1	*	Mid-Point of Line B2-B7		-1	0	0	+1
    Mid-Point of Line B3-B5		+1	0	0	+1	*	Mid-Point of Line B4-B6		-1	0	0	-1
    Mid-Point of Line B3-B6		0	+1	-1	0	*	Mid-Point of Line B4-B5		0	-1	+1	0
    Mid-Point of Line B3-B7		0	+1	0	+1	*	Mid-Point of Line B4-B8		0	-1	0	-1
    Mid-Point of Line B3-B8		+1	0	-1	0	*	Mid-Point of Line B4-B7		-1	0	+1	0
    Mid-Point of Line B5-B7		0	0	+1	+1	*	Mid-Point of Line B6-B8		0	0	-1	-1
    Mid-Point of Line B5-B8		+1	-1	0	0	*	Mid-Point of Line B6-B7		-1	+1	0	0

  And we see here that :

  • the 24 mid-points arising from the { Series-A Vectors }

    and

  • the 24 mid-points arising from the { Series-B Vectors }

    do match on a one-to-one basis ;

  and in particular :

  • the [ mid-point of Line A1-A3 ] and the [ mid-point of Line B1-B3 ]
    • do share the same co-ordinates ,

    so that we may now conclude that :

    • the [ 2 said lines above ] do intersect one-another at their mid-points .

  As such , the { 4 nodal-points Structure A1-B1-A3-B3 } is a square .

• Let us now proceed to identify the full set of 24 Structural Squares for the { 4-Dimension Quasi-Cube } .

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Listing-out the 24 Squares :

• Let us now list-out the 24 Squares
  • arising from the { 4-Dimension Quasi-Cube } .

    Square I.D.	* * *	Position Vector for Center-Point of the Square					* * *	Square I.D.	* * *	Position Vector for Center-Point of the Square
    	*	Vector I.D.	Co-ordinates				*		*	Vector I.D.	Co-ordinates
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A1-B1-A3-B3		Vector D1	+1	+1	0	0	*	Square A2-B2-A4-B4		Vector D2	-1	-1	0	0
    Square A5-B5-A7-B8		Vector D3	+1	-1	0	0	*	Square A6-B6-A8-B7		Vector D4	-1	+1	0	0
    Square A5-B2-A8-B3		Vector D5	0	0	-1	+1	*	Square A6-B1-A7-B4		Vector D6	0	0	+1	-1
    Square A1-B5-A4-B7		Vector D7	0	0	+1	+1	*	Square A2-B6-A3-B8		Vector D8	0	0	-1	-1
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A1-B3-A5-B5		Vector E1	+1	0	0	+1	*	Square A2-B4-A6-B6		Vector E2	-1	0	0	-1
    Square A3-B1-A7-B8		Vector E3	+1	0	0	-1	*	Square A4-B2-A8-B7		Vector E4	-1	0	0	+1
    Square A3-B3-A8-B6		Vector E5	0	+1	-1	0	*	Square A4-B4-A7-B5		Vector E6	0	-1	+1	0
    Square A1-B1-A6-B7		Vector E7	0	+1	+1	0	*	Square A2-B2-A5-B8		Vector E8	0	-1	-1	0
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A3-B3-A5-B8		Vector F1	+1	0	-1	0	*	Square A4-B4-A6-B7		Vector F2	-1	0	+1	0
    Square A1-B1-A7-B5		Vector F3	+1	0	+1	0	*	Square A2-B2-A8-B6		Vector F4	-1	0	-1	0
    Square A1-B3-A8-B7		Vector F5	0	+1	0	+1	*	Square A2-B4-A7-B8		Vector F6	0	-1	0	-1
    Square A3-B1-A6-B6		Vector F7	0	+1	0	-1	*	Square A4-B2-A5-B5		Vector F8	0	-1	0	+1

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Some Features associated with the 24 Squares :

• Let us now take a look at some special features associated with the 24 Squares .

  FEATURE ONE :

  • Since we have , as identified above , 32 pairs of Position Vectors intersecting at an angle of [ 60 degrees ] ,
    • we have indeed only 32 lines of { Length [ 2 ] } to construct the 24 Squares .

    Since each square is 4-sided and will need 4 lines to construct ,

    • 24 squares will require 96 lines of { Length [ 2 ] } .

    As such , each of the 32 lines of { Length [ 2 ] } will need to be shared by 3 squares ,

    • i.e. : [ 96 ] = 3 x [ 32 ] .

  FEATURE TWO :

  • The opposite sides of a square are parallel so that :
    • the square is always made up of 2 sets of { 2 parallel lines } .

    A special feature of the 24 Squares here is that :

    • for the 32 lines of { Length [ 2 ] } :
      • each { set of 8 lines } will be parallel to one (1) of the { 4 Axial Directions } .

    • for the 24 squares :
      • we take note here that :
        • pairing any two (2) of the { 4 Axial Directions } will form a plane

          and

          pairing the two (2) remaining { Axial Directions } will form a second plane orthogonal to the first ;

        so that :

        • the total number of planes arising from pairing the { 4 Axial Directions } is six (6) .

        As such :

        • for each of the six (6) planes thus identified , we will have 4 Squares residing in planes parallel to it ;
          • i.e. : [ 24 ] = 6 x [ 4 ] .

• We shall have more comments on the structural positioning of the 24 Squares ,
  • later in Section XVII ,
    • after we have done the { Quasi-Octahedron in 4-Dimensions } in Section XVI .

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What's Next :

• We notice here that :
  • we can construct 12 lines joining the center-points of each pair of { opposite squares } ,
    • and 12 lines are most-likely symmetric .

  Let us take a look at these 12 possibly-symmetric lines in 4-Dimensions , next .

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go to the next Section : Section IX - 12 Symmetric Lines in a { 4-D Vector Space }

go to the last Section : Section VII - Setting up the 8 Cross-Diagonal Lines of the { 4-D Quasi-Cube }

return to the Symmetry Through-put HomePage

Original dated 2017-7-23